There is a MUCH easier way to do this. Cube both sides of the original equation. This gives x^3 + 1/x^3 =0 Cube both sides again This gives x^9 + 1/x^9 = 0 Multiply both sides by x^8 +1/x^8 This gives x^17 + 1/x^17 +x +1/x =0 Thus x^17 + 1/x^17 = - sqrt(3)
If we square (x+1/x) 3 times we get (x⁸+1/x⁸)=-1. We also cube it 2 times and get (x⁹+1/x⁹)=0. We multiply both expressions and the result is x¹⁷+1/x¹⁷=-√3.
Another approach The original equation is: x+1/x=√3 By squaring this equation and simplifying we get: x^2+1/x^2 =1 Therefore: x^4-x^2+1=0 ⟹ x^6=-1 ⟹ x^18=-1 ⟹ x^17=-1/x Therefore: x^17+1/x^17 =-1/x-x=-(x+1/x)=-√3
Syber's second method requires the leap of insight that sqrt(3) = 2*cos(pi/6). What if we just solve the original equation, a quadratic? We find x = (sqrt(3) +/- i)/2 = e^(i*pi/6), and the rest follows naturally.
I used a third method, in which I found an expression for x^2, x^4, x^8, and x^16. Finally, I plugged in the expression for x^17 into the equation and set common denominators from there.
This explanation is too unclear. Please be more accurate. We All need to learn and we Are not in the same level of understanding. Sorry for being this honest. thanks for your work
There is a MUCH easier way to do this.
Cube both sides of the original equation.
This gives x^3 + 1/x^3 =0
Cube both sides again
This gives x^9 + 1/x^9 = 0
Multiply both sides by x^8 +1/x^8
This gives x^17 + 1/x^17 +x +1/x =0
Thus x^17 + 1/x^17 = - sqrt(3)
nice!
If we square (x+1/x) 3 times we get (x⁸+1/x⁸)=-1. We also cube it 2 times and get (x⁹+1/x⁹)=0. We multiply both expressions and the result is x¹⁷+1/x¹⁷=-√3.
Another approach
The original equation is:
x+1/x=√3
By squaring this equation and simplifying we get:
x^2+1/x^2 =1
Therefore:
x^4-x^2+1=0 ⟹ x^6=-1 ⟹ x^18=-1 ⟹ x^17=-1/x
Therefore:
x^17+1/x^17 =-1/x-x=-(x+1/x)=-√3
Syber's second method requires the leap of insight that sqrt(3) = 2*cos(pi/6). What if we just solve the original equation, a quadratic? We find x = (sqrt(3) +/- i)/2 = e^(i*pi/6), and the rest follows naturally.
Cubing both sides gives a neat result x^6=-1. x^17=(x^6)^3/x.
I used a third method, in which I found an expression for x^2, x^4, x^8, and x^16. Finally, I plugged in the expression for x^17 into the equation and set common denominators from there.
Made it complicated unessoryily.(.x^8+1/x^8)× (x^9+1/x^9) runs smooth
Another approach could be doubling the power of x:
x² = √ 3 x - 1
x⁴ = (x²)² = √ 3 x - 2
x⁸ = (x⁴)² = 1 - √ 3 x
x^16 = (x⁸)² = √ 3 x - 2
x^17 = √ 3 x²- 2x = x - √ 3.
x^17 + 1/x^17
= x - √ 3 + 1/(x - √ 3) = -√ 3.
perfect❤❤❤
😊🎉😊👍👍👍😊🎉😊 glad your voice is coming back
x=e^(-iπ/6)...?=2cos(17π/6)=-2cosπ/6=-√3...mah,dovrei ricontrollare
x + 1/x = √3
x² + 1/x² = 1
(x² + 1/x²)(x + 1/x) = x³ + 1/x³ + x + 1/x
√3 = x³ + 1/x³ + √3
x³ + 1/x³ = 0
x⁶ = -1
x¹⁷ = (x⁶)³/x = -1/x
1/x¹⁷ = -x
*x¹⁷ + 1/x¹⁷ = -(x + 1/x) = -√3*
super
Let f(x) = x +1/x = sqrt(3) [f(x)]^2 =3 implies f(x^2)=1 f(x)*f(x^2) = sqrt(3) implies f(x^3)=0 f(x^3)*f(x^6) = f(x^9) +f(x^3) =0 implies f(x^9)=0
f(x^9)*f(x^8) = f(x^17) + f(x) thus f(x^17) = f(x^9)*f(x^8) - f(x) = 0*f(x^8) -sqrt(3) = -sqrt(3)
(x + 1/x)^3 =x^3 +1/x^3 + 3x*1/x*(x + 1/x)
3√3 = x^3 + 1/x^3 + 3*√3 => x^3 + 1/x^3 = 0 => x^6 = -1
x^17 = x^18 / x = (-1)^3 / x = -1/x => 1/x^17 = -x
x^17 + 1/x^17 = -1/x - x = -(x + 1/x) = -√3
your voice was depressive recent day
x=½(√3]±i)=e^±⅓πi
?=e^⅓πi+e^-⅓πi=1
x^2+1/x^2=1 x^4-x^2+1=0 (x^2+1)(x^4-x^2+1)=0=(x^6+1) x^6=-1 x^6=-1 x^17+1/x^17=x^18/x+x/x^18=-1/x-x=-(x+1/x)=-sqrt(3)
This explanation is too unclear. Please be more accurate. We All need to learn and we Are not in the same level of understanding. Sorry for being this honest. thanks for your work
@@Marcus-y1m x + 1/x = √3
x² + 1/x² = 1
(x² + 1/x²)(x + 1/x) = x³ + 1/x³ + x + 1/x
√3 = x³ + 1/x³ + √3
x³ + 1/x³ = 0
x⁶ = -1
x¹⁷ = (x⁶)³/x = -1/x
1/x¹⁷ = -x
*x¹⁷ + 1/x¹⁷ = -(x + 1/x) = -√3*
@@Marcus-y1m x + 1/x = √3
x² + 1/x² = 1
(x² + 1/x²)(x + 1/x) = x³ + 1/x³ + x + 1/x
√3 = x³ + 1/x³ + √3
x³ + 1/x³ = 0
x⁶ = -1
x¹⁷ = (x⁶)³/x = -1/x
1/x¹⁷ = -x
*x¹⁷ + 1/x¹⁷ = -(x + 1/x) = -√3*
@@SidneiMV wow, it was perfect!
Thanks for helping. Now i got it.